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Rudik [331]
4 years ago
9

You shine light with a wavelength of 400 nm at two small slits (I and II) and observe an interference pattern on a screen. What

is the difference in path lengths between the light that travels from slit I to the first dark fringe of the pattern and the light that travels from slit II to the same dark fringe?
Physics
1 answer:
GuDViN [60]4 years ago
3 0

Answer:

2 x 10⁻⁷ m

Explanation:

Since the light from two slits interfere to form dark fringe on the screen ,

the path difference must be odd multiple of λ / 2 .

Since it is the first dark fringe the factor of odd multiple must be one.

So path difference must be

( 2n +1 ) x λ / 2 .

for first dark fringe

n = 0

path difference  =  λ / 2

400 x 10⁻⁹ / 2 m

= 2 x 10⁻⁷ m

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siniylev [52]

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

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M=mass

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K.E =0.5*55*0.6^2

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Explanation:

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g100num [7]

Answer:

0.3956

Explanation:

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3 years ago
An automobile tire is rated to last for 35,000 miles. to an order of magnitude, through how many revolutions will it turn?
Furkat [3]
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7 0
3 years ago
I dont know how to do any of this so someone please help (the way it was solved has to be present for each thing) i will give br
Alexus [3.1K]
I cannot see all the questions, what is 18,19 and 21? (:
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A newly discovered planet is found to have a density if 2/3pe and a radius of 2RE, where PE and RED are the density and radius o
Liono4ka [1.6K]
Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"

Solution:
The gravitational acceleration at Earth's surface is given by:
g= \frac{GM}{r^2} 
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius

The Earth's mass can be rewritten also as the product between the Earth's density, d, and its volume (the volume of a sphere of radius r):
M=dV=d ( \frac{4}{3} \pi r^3)=  \frac{4}{3} \pi d r^3 

Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
g' =  \frac{GM'}{r'^2} (1)
so we need to find M' and r'.

The problem says the radius of the new planet is twice the Earth's radius: 
r'=2r (2)
and that its density is 2/3 of Earth's density:
d'= \frac{2}{3} d
so the mass M' of the new planet is, with respect to the Earth's mass:
M' = d'V' = \frac{4}{3} \pi d' (r')^3 =  \frac{4}{3} \pi ( \frac{2}{3}d) (2r)^3 = ( \frac{4}{3} \pi d r^3 )( \frac{16}{3}) =  \frac{16}{3} M (3)

And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
g'= \frac{G( \frac{16}{3}M) }{(2r)^2}=  \frac{GM}{r^2}  \frac{4}{3} =  \frac{4}{3}g
And since g=9.81 m/s^2, we find
g'=  \frac{4}{3}(9.81 m/s^2)=13.1 m/s^2
8 0
4 years ago
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